Question

ii) Point E is in the exterior of a circle. A secant through E intersects the circle at points A and B and a tangent through E touches the circle at point T then prove that EA X EB = ET^2​

Answer 1

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Answer 2

ii) Point E is in the exterior of a circle. A secant through E intersects the circle at points A and B

Given,

Point E is in the exterior of a circle.

A secant through E intersects the circle at points A and B.

Now consider,

In ∆ EAT and ∆ ETB,

∠ ETA ≅ ∠ EBT  (using Theorem of angle between tangent and secant)  

∠ AET ≅ ∠ TEB  (common angle)

∴ ∆EAT ~ ∆ETB      ( using AA criteria )

⇒  EA / ET = ET / EB   ( c.p.c.t )  

∴ EA × EB = ET^2

Hence the proof.